on a numpy 2D array - how to set last N array elements in each row to zero when N changes over rows

Question

Say, I have a 2D numpy array consists of 20 elements, for example:

arr = np.array([[1, 2, 15, 7],[9, 11, 17, 19],[5, 7, 5, 8],[19, 4, 1, 45],[10, 7, 14, 8]])

and an additional array:

to_zero = np.array([0, 2, 1, 3, 2])

now, for each row i I would like to make the last to_zero[i] elements equal to zero, so eventually we will get the following result:

res = np.array([[1, 2, 15, 7],[9, 11, 0, 0],[5, 7, 5, 0],[19, 0, 0, 0],[10, 7, 0, 0]])

I would like to do this operation on a very big array. Is there any way to do this operation vectorized, with no loops, and no auxiliary arrays?

Possible duplicate of How to select elements in numpy array with given starting point indices — Brenlla
– Brenlla, Commented Oct 2, 2019 at 13:55

Divakar · Accepted Answer · 2019-10-02 14:04:16Z

5

Use broadcasted-comparison to get a mask of those trailing ones and then mask the input -

In [63]: r = np.arange(arr.shape[1])[::-1]

In [66]: mask = to_zero[:,None]>r

In [69]: mask # mask of trailing places to be reset in input
Out[69]: 
array([[False, False, False, False],
       [False, False,  True,  True],
       [False, False, False,  True],
       [False,  True,  True,  True],
       [False, False,  True,  True]])

In [67]: arr[mask] = 0

In [68]: arr
Out[68]: 
array([[ 1,  2, 15,  7],
       [ 9, 11,  0,  0],
       [ 5,  7,  5,  0],
       [19,  0,  0,  0],
       [10,  7,  0,  0]])

Alternative to getting r would be with np.arange(arr.shape[1]-1,-1,-1).

Alternative to get the final output with element-wise multiplication : arr*~mask.

Or construct the flipped-mask with the flipped-comparison and then multiply -

In [75]: arr*(to_zero[:,None]<=np.arange(arr.shape[1]-1,-1,-1))
Out[75]: 
array([[ 1,  2, 15,  7],
       [ 9, 11,  0,  0],
       [ 5,  7,  5,  0],
       [19,  0,  0,  0],
       [10,  7,  0,  0]])

For large arrays, leverage multi-cores with numexpr -

In [78]: import numexpr as ne

In [79]: ne.evaluate('arr*mask',{'mask':to_zero[:,None]<=np.arange(arr.shape[1]-1,-1,-1)})
Out[79]: 
array([[ 1,  2, 15,  7],
       [ 9, 11,  0,  0],
       [ 5,  7,  5,  0],
       [19,  0,  0,  0],
       [10,  7,  0,  0]])

edited Oct 2, 2019 at 14:04

answered Oct 2, 2019 at 13:50

Divakar

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3 Comments

aviyaChe Over a year ago

Is there any way to do this operation vectorized, with no auxiliary arrays as noted in the question?

Divakar Over a year ago

@aviyaChe Basically you are asking for a builtin and sadly there isn't any in NumPy. Also, we are constructing a mask/boolean array that occupies 1/8th of an int array on unix systems.

hpaulj Over a year ago

Even though the selection for a given row is a slice, the selection for the whole array requires some form of advanced indexing, identifying each element. That's what the mask does.

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